chapter 8 electrochemical equilibria
8.1Introduction………..………………………………………………………………….….1
8.2Redox Equilibria and Electron Activity………………………………………………..2
8.2.1Redox Reactions…………………………………………………………………..2
8.2.2Electron Activity…………………………………………………………………..2
8.3Electrochemical Cells and Electrode Potentials………………………………………..7
8.3.1Metal Electrodes…………………………………………………………………..7
8.3.2Electrochemical Cells and Redox Equilibria.…………………………………….8
8.3.3Standard Electrode Potentials…...... …………………….12
8.3.4Electron Activity and Redox Potential…………………………………………... 15
8.3.5Minimum Electrolyzing Voltage………………………………………………...17
8.4Electrochemical Selectivity…………………………………………………………………………. 23
8.4.1Selectivity of the Anodic Process………………………………………………………….23
8.4.2Selectivity of the Cathodic Process………………………………………………………..26
8.4.3Electrodeposition Reactions………………………………………………………………… 27
8.4.4Electrorefining Reactions……………………………………………………………………35
8.5Semiconductor Electrodes……………………………………………………………...38
8.5.1Energy Levels of Redox Couples………………………………………………... 38
8.5.2Energy Levels and Electron Transfer……………………………………………. 39
8.1 Introduction
The two broad categories of chemical reactions are atom transfer reactions and electron transfer reactions. In atom transfers, chemical reaction does not lead to changes in the oxidation states of the constituent atoms of the reactants. In Chapters 5-7 we limited our discussion to such reactions. In the present chapter, we turn our attention to the other major class of reactions, i.e., the case where chemical reaction introduces changes in oxidation states. While electron transfer reactions can occur in homogeneous solution, they oftentimes take place in heterogeneous systems. Such interfacial electron transfer reactions form the basis of important industrial processes such as electroplating, electrowinning, and electrorefining. The term electrochemical may be used strictly to refer only to electron transfer reactions at interfaces. However, a more flexible usage of the term is also widely accepted, where it refers to both homogeneous and heterogeneous processes and to the transfer of positive as well as negative charges. In this chapter we shall adopt this more general usage.
8.2Redox Equilibria and Electron Activity
8.2.1Redox Reactions
Consider the deposition of metallic copper via reaction of cupric ion with hydrogen gas:
Cu2+ + H2(g) = Cu + 2H+(8.1)
This reaction can be viewed as consisting of two steps:
H2(g) = 2H+ +2e-(8.2)
Cu2+ + 2e- = Cu(8.3)
That is, the copper deposition reaction involves the transfer of two electrons from a hydrogen gas molecule to a cupric ion. Reactions which involve such electron transfers are called oxidation-reduction or redox reactions.
The species which releases electrons, i.e., the electron donor is called the reductant, while the electron acceptor is called the oxidant. Another way of looking at this process is to think of a reductant as an electron complex. Thus an oxidation reaction is one in which an electron complex or reductant decomposes, i.e., releases its electron, while a reduction reaction is one in which an electron complex or a reductant is formed. Thus in reaction 8.1, Cu and H2 are reductants, since according to Equations 8.2 and 8.3, they are electron complexes. On the other hand, H+ and Cu2+ are oxidants. Furthermore, Equations 8.2 and 8.3 are respectively oxidation and reduction reactions.
8.2.2Electron Activity
It is convenient to define a function p which is a measure of electron activity (cf, pH, a measure of proton activity):
p = -log{e-}(8.4)
When p is large and positive, (i.e., low electron activity, {e}), we have strong oxidizing conditions. Since oxidants are electron acceptors, a system containing a high concentration of oxidant must necessarily contain a small amount of "free" electrons, i.e., the system will be characterized by a low level of electron activity. When p is small or negative (i.e., high electron activity) we have strongly reducing conditions.
Consider the general redox reaction,
aA + bB + ne- = cC + dD(8.6)
The corresponding reaction quotient, Q, is given by:
(8.7)
where
Q' = (8.7a)
At equilibrium,
logQeq = logK(8.8)
Therefore,
If each of the species A, B, C, and D is present at unit activity, it follows that
logK = (p)eq, A, B, C, D at unit activity(8.10)
In order to provide a less clumsy way of writing Equation 8.10, the quantity p° is defined, where
p° = logK(8.11)
Therefore, Equation 8.9 can be rewritten as:
where the subscript "eq" has been omitted for convenience.
Example 8.1 Redox Equilibria
Calculate p values for the following equilibrium systems (assume that the solutions are infinitely dilute):
(a)An acidic solution 10-3M in V3+ and 10-5M in V2+
(b)A solution at pH 2 in equilibrium with hydrogen gas at 0.5 atm.
(c)A solution containing 10-5 mol/L RuO42- in equilibrium with RuO22H2O(s) at pH 14.
(d)A solution containing 10-4 mol/L Fe2+ in equilibrium with Fe3O4 at pH 6.
(e)An acidic solution containing 10-5 mol/L Cd2+ and at equilibrium with CdS(s) and elemental sulfur.
The following data are available:
V3+ + e- = V2+logK = -4.31(1)
2H+ + 2e- = H2(g)logK = 0(2)
RuO42- + 4H+ + 2e- = RuO22H2O(s)logK = 67.33 (3)
Fe3O4(s) + 8H+ + 2e- = 3Fe2+ + 4H2OlogK = 36.81(4)
Cd2+ + S(s) + 2e- = CdS(s)logK = 12.16(5)
Solution
(a)It follows from Equation 1 that
Q = [V2+]/[V3+] {e-}
At equilibrium log Q = log K and therefore
logK = log ([V2+]/[V3+]) - log {e-}
Thus,
p = logK - log ([V2+]/[V3+])
= (-4.31) - log(10-5/10-3) = (-4.31) - (log 10-2)
= (-4.31) - (-2.0) = -2.31
Alternatively we can use Equation 8.12 directly:
p = po - (1/n) log Q'(8.12)
Comparison of Equations 1 and 8.6 indicates that n= 1. Therefore from Equation 8.11,
po = (1/n) logK = -4.31(6)
Also comparison of Equations 1 and 8.6 in the light of Equation 8.7a gives:
logQ' = log ([V2+]/[V3+]) = log (10-5/10-3) = - 2.0(7)
It follows from Equations 6, 7, and 8.12 that
p = -4.31 - (-2.0) = -2.31
(b)Referring to Equation 2,
Q' = PH2/{H+}2
Thus,
log Q' = log PH2 + 2pH = log (0.5) + 2(2) = (-0.30) + 4 = -3.70
Also n = 2, and log K = 0. Therefore from Equation 8.11, po = 0. Therefore by using Equation 8.12,
p = (0) - (1/2) (-3.70) = 1.85
(c)From Equation 3,
logQ' = - log[RuO4 2-] + 4pH = -log(10-5) + 4(14) = 5 + 56 = 61
Also,
po = (1/n)logK = (1/2) (67.33) = 33.67
It follows from Equation 8.12 that:
p = (33.67) - (1/2) (61) = 3.17
(d)From Equation 4,
logQ' = 3log[Fe2+] + 8pH = 3log(10-4) + 8 (6) = -12 + 48 = 36
Also,
po = (1/n) logK = (1/2) (36.81) = 18.4
Thus it follows from Equation 8.12 that
p = (18.4) - (1/2) (36) = 0.4
(e)From Equation 5,
logQ' = -log[Cd2+] = - log(10-5) = 5
po = (1/2) (12.16) = 6.08
Therefore
p = (6.08) - (1/2) (5) = 6.08 -2.5 = 3.58
example 8.2 Graphical Representation of Redox Equilibria
Consider the reaction
V3+ + e- = V2+ logK = -4.31(1)
Under equilibrium conditions, we have (using concentrations in place of activities),
K = (2)
Also there is the following mass balance relation for the total dissolved vanadium, [V]T:
[V]T = [V2+] + [V3+](3)
It follows from Equations 2 and 3 that:
[V3+] = (4)
and
[V2+] = (5)
Thus,
log[V3+] = log[V]T - logK - log[{e-} + K-1](6)
log[V2+] = log[V]T - p - log[{e-} + K-1 ](7)
These results can be depicted graphically. We can make use of the following asymptotic conditions: When {e-}>K-1, i.e., p<po, Equations 6 and 7 reduce to Equations 8 and 9 respectively:
Figure E8.2 log[V] versus pplot for V3+ and V2+ when [V]T = 10-4M.
log[V3+ ] = log[V]T - po + p
log[V2+] = log[V]T(9)
On the other hand, when {e-} < 1/K, i.e., p> po, Equations 6 and 7 become respectively:
log[V3+] = log[V]T(10)
log[V2+] = log[V]T + po - p(11)
Figure E8.2 shows a plot of log[V] versus p for V3+ and V2+ when [V]T = 10-4M.
8.3Electrochemical Cells and Electrode Potentials
8.3.1Metal Electrodes
When a piece of metal is immersed in an electrolyte, the metal is called an electrode. The dissolution of the metal is described by Equation 8.13:
M = Mz+ + ze-(8.13)
If the electrons are more "soluble" than the metal ion, the electric charge on the electrode surface will be positive, as shown by Figure 8.1a. That is, the electrode/solution interface will acquire a positive galvanic potential. On the other hand, if the metal ion is more soluble that the electrons, the electrode surface will be negatively charged as shown in Figure 8.2b, and the electrode/solution interface will be characterized by a negative galvanic potential.
Figure 8.1. Metal electrodes and the charging of the electrode/solution interface:
(a) Positive electrode surface; (b) Negative electrode surface.
Figure 8.2. Convention for anodes and cathodes
Electrodes are termed anodes and cathodes depending on the direction of flow of electrical charges, as illustrated in Figure 8.2. When negative charge flows from the electrolyte to the electrode, an anodic current is said to be flowing and the electrode is called an anode. On the other hand, when electrons leave the electrode and pass into the electrolyte, a cathodic current is said to be flowing and the electrode is designated the cathode. The anode is where oxidation occurs, and an anodic current gives rise to an anodic reaction, i.e., oxidation. Correspondingly, the cathode is where reduction occurs, and a cathodic current gives rise to a cathodic reaction, i.e., reduction. It is important to note that an anode or cathode is defined strictly by the direction of electron flow. This definition has nothing to do with the sign of the electrode potential.
8.3.2Electrochemical Cells and Redox Equilibria
When the electrolytes of two electrodes have a common boundary, the arrangement is called an electrochemical cell. Electrochemical cells may be galvanic or electrolytic. In galvanic cells reactions occur spontaneously, giving rise to electrochemical potentials. On the other hand, in the case of electrolytic cells, electrochemical reaction results from externally applied potential. Electrochemical cells are often represented by cell diagrams. For example, a cell consisting of two metal electrodes (M1 and M2) can be depicted schematically as:
M1|M| |M|M2
where the symbols (|) represent the respective metal/electrolyte interfaces and the symbol (| |) depicts the common boundary of the two electrolytes containing the ions M1z1+ and M2z2+ respectively.
Consider the reaction
aA + bB = cC + dD(8.14)
where the free energy of reaction is given by,
G = cC + dD - aA - bB(8.15)
Suppose this reaction can proceed electrochemically in a galvanic cell such that the partial reactions occurring at the anode and cathode may be described with Equations 8.16 and 8.17 respectively:
aA = cC + ze(8.16)
bB + ze= dD(8.17)
where eand erepresent the electrons associated with the anode and cathode respectively. An overall reaction may be obtained by combining Equations 8.16 and 8.17:
aA + bB + ze= cC + dD + ze(8.18)
The free energy change associated with Equation 8.18 is given by
G' = cC + dD - aA - bB + z(ea - ec)(8.19)
At equilibrium, G' = 0 and therefore Equation 5.19 becomes:
z(ea -ec) = - (cC + dD - aA - bB)(8.20)
It can be seen from a comparison of Equations 8.15 and 8.20 that
z(ea - ec) = -G(8.21)
Now, the chemical potential of an electron at a given location is the reversible work required to transfer one mole of electrons from infinity to that location and it is related to the electrical potential, , at that location as,
e = zeF = -F(8.22)
where ze = -1 is the charge on the electron, and F is the Faraday constant with a value of 96487.0 coulombs mol-1 (i.e., joul. volt-1 mol-1 or 23060.9 cal volt-1 mol-1. Thus introducing Equation 8.22 into Equation 8.21 gives
zF(c - a) = -G(8.23)
where a and c are the electrical potentials of the anode and cathode respectively.
The overall reaction is spontaneous for a galvanic cell, i.e., G is negative. It follows from Equation 8.23 then that the cell potential, defined by Equation 8.24, is positive for a galvanic cell:
Ecell = c - a = -G/zF > 0 (8.24)
Thus according to Equation 8.24, in a galvanic cell the more positive electrode is the cathode. On the other hand, for an electrolytic cell the overall reaction is not spontaneous, i.e., G is positive and therefore Ecell = c - a < 0; that is, in this case the anode is the more positive electrode. Therefore, the terminals of electrochemical cells are classified as plus or minus, as shown in Figure 8.3. Note that electrons always leave a cell, whether galvanic or electrolytic, at the anode. On the other hand, electrons always enter a cell at the cathode.
Figure 8.3 Sign convention for terminals of electrochemical cells.
According to Equations 8.23 and 8.24,
Ecell = c - a = -G/zF(8.25)
Thus recalling for the overall reaction that G = Go + RT ln Q, it follows that
Ecell = (-Go/zF) - (2.303 RT/zF) log Q(8.26)
It should be noted that G, Go, and Q refer to the overall reaction, i.e., Equation 8.14. Recalling further, that Go = - 2.303 RT log K, Equation 8.26 can be rewritten as:
Ecell = (2.303 RT/zF) logK - ) log Q(8.27)
or
Ecell = Eo - (2.303 RT/zF) log Q(8.28)
where Eo is the standard EMF of the cell and is given by:
Eo = -Go/zF = (2.303 RT/zF) log K (8.29)
It can be seen by comparing Equations 8.28 and 8.29 that the standard EMF of the cell (Eo) represents the equilibrium cell potential under conditions where log Q has a value of unity. The definition of Eo is based on the specific case where all the aqueous species have unit activities and gases are at 1 atm pressure, and solid phases are in their most stable forms.
example 8.3 The Standard EMF of the Daniel Cell
The Daniel cell is a galvanic cell based on a zinc anode and a copper cathode. The cell diagram may be represented as:
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
(a) Write down the respective half-cell reactions
(b) Write down the overall cell reaction
(c) Determine the standard cell potential
The following data are available: G= -35.18 kcal/mole, G= 15.6 kcal/mole.
Solution
(a)Since the zinc electrode is the anode, it must undergo an oxidation reaction, i.e., it must release electrons:
Zn(s) = Zn2+(aq) + 2e-(1)
On the other hand, the copper electrode is the cathode and therefore it is the site of the corresponding reduction reaction, i.e., a reaction involving the consumption of electrons:
Cu2+(aq) + 2e- = Cu(s) (2)
(b) The overall reaction is obtained by combining Equation 1 and 2:
Zn(s) + Cu2+(aq) = Zn2+(aq) + Cu(s)(3)
(c)The standard cell potential is given by Equation 8.29:
Eo = -Go/zF = (2.303RT/zF) log K(8.29)
It follows from Equation 3 that:
Go = G[Zn2+(aq)] + G[Cu(s)] - G[Zn(s)] - G[Cu2+(aq)]
= (-35.18) + (0) - (0) - (15.6) = - 50.78 kcal/mole
Thus from Equation 8.29, with z = 2 (see Equations 1 and 2):
Eo = -(-50.78 kcal/mole)/2(23.0609 kcal V-1 mole-1) = 1.10 V.
8.3.3 Standard Electrode Potentials
The potential of an electrode does not have an absolute value, rather it is specified relative to the potential of a reference electrode. A common reference electrode is the standard hydrogen electrode (S.H.E.), which consists of a platinum electrode immersed in an aqueous solution of unit hydrogen ion activity and in contact with hydrogen gas at one atmosphere pressure. The standard hydrogen electrode is arbitrarily assigned a zero potential.
An electrochemical cell consisting of a metal electrode and the standard hydrogen electrode can be represented schematically as:
Pt, H2 (PH2 = 1 atm)| H+({H+} = 1) | | Mz+ ({Mz+})|M
The difference between the potential of the metal electrode and that of the standard hydrogen electrode is termed the electromotive force or EMF of the cell. The cell EMF is given by
Ecell = Eel - Eref = EM/Mz+ - EH2/H+ (8.30)
By definition, when PH2 = 1 atm and {H+} = 1, EH2/H+ = 0. Therefore, under these circumstances,
Ecell = EhM/Mz+(8.31)
where the symbol "h" is used in Eh to signify that the potential is measured with reference to the standard hydrogen electrode. Thus it follows from Equation 8.31 that the electrode potential of a metal is the EMF of a cell formed between the metal electrode and the standard hydrogen electrode.
The half-cell reactions corresponding to the above electrochemical cell can be written for the metal and hydrogen electrodes respectively as:
Mz+ + ze- = M KM(8.32)
zH+ + ze- = (z/2)H2(g) KH(8.33)
The overall reaction is
Mz+ + (z/2) H2(g) = M + zH+(8.34)
Thus invoking Equation 8.28, the corresponding cell potential is given by
Ecell = Eo - (2.303RT/zF) [zlog{H+} - log{Mz+} - (z/2) log PH2](8.35)
Therefore, when {H+} = 1.0 and PH2 = 1 atm,
Ecell = EhM/Mz+ = Eh+ (2.303 RT/zF) log{Mz+}(8.36)
Furthermore, when {Mz+} = 1.0,
Ecell = Eh= -Go/zF = (2.303 RT/zF) log K(8.37)
where EhoM/Mz+ is termed the standard electrode potential or the standard reduction potential for the M/Mz+ electrode. Table 8.1 presents values of standard electrode potentials for selected metals.
Metal electrodes with negative Eho values tend to behave as depicted in Figure 8.1b; that is, they will dissolve according to
M(electrode) = Mz+ (aq) + ze- (electrode)(8.38)
Thus, when electrical contact is made between the metal electrode and the Pt of the hydrogen electrode, electrons will flow to the platinum and result in the evolution of hydrogen:
2H+ + 2e- = H2(g)(8.39)
In Figure 8.4 electrons leave the cell through the M electrode and therefore this electrode represents the anode of the cell, while the hydrogen electrode is the cathode.
Table 8.1. Standard Reduction Potentials (Eho) in Acid Solution*
CoupleEoVolts
Li+ + e- = Li -3.045
K+ + e- = K -2.925
Rb+ + e- = Rb -2.925
Ba2+ + 2e- = Ba -2.90
Sr2+ + 2e- = Sr -2.89
Ca2+ + 2e- = Ca-2.87
CoupleEoVolts
Na+ + e- = Na -2.174
La3+ + 3e- = La -2.52
Mg2+ + 2e- = Mg -2.37
Sc3+ + 3e- = Sc -2.08
Th4+ + 4e- = Th -1.90
Be2+ + 2e- = Be -1.85
Hf4+ + 4e- = Hf -1.70
Al3+ + 3e- = Al -1.66
Ti2+ + 2e- = Ti -1.63
Zr4+ + 4e- = Zr -1.53
U4+ + 4e- = U -1.50
______
Mn2+ + 2e- = Mn -1.18
Nb3+ + 3e- = Nb -1.1
Zn2+ + 2e- = Zn -0.763
Cr3+ + 3e- = Cr -0.74
Ga3+ + 3e- = Ga -0.53
Fe2+ + 2e- = Fe -0.44
Cd2+ + 2e- Cd -0.403
In3+ + 3e- = In -0.342
Co2+ + 2e- = Co -0.277
Ni2+ + 2e- = Ni -0.250
Mo3+ + 3e- = Mo ~-0.20
Sn2+ + 2e- = Sn -0.136
Pb2+ + 2e- = Pb -0.126
2H++ 2e- = H2 -0.00
Cu2+ + 2e- = Cu +0.337
Hg+ 2e- = 2Hg +0.789
Ag+ + e- = Ag +0.7991
Rh3+ + 3e- = Rh ~ + 0.8
Pd2+ + 2e- = Pd +0.987
Au3+ + 3e- = Au +1.50
*Source: W. M. Latimer, Oxidation Potentials
Figure 8.4 Behavior of metal electrodes with negative standard
electrode potentials (Eho): evolution of hydrogen gas.
8.3.4Electron Activity and Redox Potential
According to convention the equilibrium constant for the hydrogen reaction (Equation 8.33) is unity, i.e., KH = 1. Thus, the equilibrium constant for the overall reaction (Equation 8.34) is given by
logK = logKM - logKH = logKM(8.40)
It follows therefore from Equations 8.36 to 8.40 that
In view of the previous discussion in Section 8.2, Equation 8.32 can also be considered in terms of electron activity, i.e.,
p = (1/z) logKM + (1/z) log {Mz+}(8.42)
where po = (1/z) logKM. A comparison of Equations 8.41 and 8.42 reveals the following relationships between Eh and p:
Eh = (2.303 RT/F) p(8.43)
Eho = (2.303 RT/F) po(8.44)
example 8.4 Standard Reduction Potentials
For each of the equilibrium systems listed in Ex. 8.1, determine the standard reduction potential as well as the Eh for the indicated solution conditions.
Solution
According to Equations 8.43, and 8.44,
Eh = (2.303RT/F) p(8.43)
Eho = (2.303RT/F) po(8.44)
(a)V3+ + e- = V2+logK = -4.31
(i)From Ex. 8.1, for logK = -4.31, p° = -4.31. Therefore Eh° = 0.059 p°,
volt = (0.059)(-4.31) V = -0.25 V
(ii)For 10-3M V3+, 10-5M V2+, p = -2.31 (Ex.8.1). Thus Eh = 0.059 p, volt
= (0.059)(-2.31)V = -0.136 V.
(b)2H+ + 2e- = H2(g)logK = 0
(i)From Ex. 8.1, for logK= 0, p° = 0. Therefore Eh° = 0V
(ii)For pH = 2, PH2 = 0.5 atm, p = 1.85. Thus Eh = 0.059p, volt =
(0.059)(1.85)V = 0.109V.
(c)RuO42- + 4H+ + 2e- = RuO2.2H2O(s)logK=67.33
(i)From Ex. 8.1, for logK = 67.33, p° = 33.67. Thus Eh° = 0.059 p° =
(0.059)(33.67)V = 1.99V.
(ii)For [RuO42-] = 10-5 mol/L and pH 14, p = 3.17 (Ex. 8.1). Thus Eh - 0.059
p = (0.059)(3.17) V = 0.187V.
(d)Fe3O4(s) + 8H+ + 2e- = 3Fe2+ + 4H2OlogK = 36.81
(i)From Ex. 8.1, p° = 18.4. Thus Eh° = 0.059 p° = (0.059)(18.4)V = 1.09V.
(ii)For [Fe2+] = 10-4 mol/L and pH 6, p = 0.4 (Ex.8.1). Thus Eh = 0.059 p
= (0.059)(0.4)V = 0.024V.
(e)Cd2+ + S(s) + 2e- = CdS(s)logK = 12.16
(i)From Ex.8.1, p° = 6.08. Thus Eh° = 0.059 p° = (0.059)(6.08)V = 0.36V.
(ii)For [Cd2+] = 10-5 mol/L, p = 3.58 (Ex.8.1). Thus Eh = 0.059 p =
(0.059)(3.58)V = 0.211V.
8.3.5Minimum Electrolyzing Voltage
Consider the reaction
Mz+ + (z/2)H2O = z/4O2 + zH+ + M(8.45)
where
G1 = z/4O2 + zH+ + M - Mz+ -(z/2)H2O(8.46)
Table 8.2 presents the Go values of this reaction for several metals. It can be seen that in all cases, the free energy change is positive and therefore unfavorable for spontaneous reaction in the forward direction. Equation 8.45 can be re-expressed in terms of two half-cell reactions, i.e., a cathodic reaction yielding a metallic phase (Equation 8.47) and an anodic reaction giving rise to oxygen evolution (Equation 8.48).
Mz+ + ze- = M(8.47
(z/2)H2O = z/4 O2 + zH+ + ze-(8.48)
Table 8.2 Values of G°r for the reactions Mz+ + (z/2) H2O = (z/4)O2 + zH+ + M
___Mz+___ / G°r (kcal/mol) / ___Mz+___ / G°r (kcal/mol)Ag+ / 9.9 / Ni2+ / 67.5
Au3+ / 20.2 / Pb2+ / 62.4
Cr3+ / 36.2 / Pd2+ / 25.3
Cu2+ / 41.0 / Sn2+ / 63.1
Fe2+ / 78.7 / Zn2+ / 91.8
The ability to achieve metal deposition via electrolysis is based on the fact that the chemical potentials of the respective electrons associated with the cathodic and anodic reactions can be altered sufficiently to reverse the sign of the free energy change corresponding to the overall reaction (Equation 8.45).
Imagine a (divided) cell within which a metal deposition process is occurring. When an external source of electrical energy is applied, the electrons at the cathode and anode acquire different chemical potentials. Thus the overall reaction (obtained by combining Equations 8.47 and 8.48) can be written as
Mz+ + (z/2) H2O + ze= z/4 O2 + zH+ + M + ze(8.49)
where subscripts c and a denote the cathode and anode respectively. Assuming that aside from the electrons, the application of the electrical potential does not change the chemical potentials of the reactants and products, it follows that the free energy change associated with Equation 8.49 can be represented as:
G2 = z/4 O2 + zH+ + M - Mz+ - (z/2)H2O + zea - zec(8.50)
Equations 8.46 and 8.50 can be combined to give
G2 = G1 + zea - zec(8.51)
In the absence of an externally applied voltage, the cathode and anode electrons have the same chemical potential, i.e.,
ea = ec(8.52)
Therefore under these circumstances,
G2 = G1(8.53)
On the other hand, at equilibrium in the presence of applied potential, G2 = 0 and therefore Equation 8.51 becomes
G1 = - z(ea - ec)eq(8.54)
But at equilibrium, the chemical potential of an electron in an electrode is the same as that of an electron in the corresponding terminal. It follows therefore that Equation 8.54 can be written as
G1 = -z(eat - ect)eq(8.55)
where the subscripts at and ct respectively represent the anode and cathode terminals.
Recalling the relationship between electrical potential and the chemical potential of an electron (Equation 8.22), we can rewrite Equation 8.55 as:
G1 = zF (at - ct)eq(8.56a)
= zFEmin(8.56b)
where Emin = (at - ct)eq is the minimum electrolyzing voltage (MEV).
Example 8.5 Minimum Electrolyzing Voltage for Copper Electrowinning
The higher the minimum electrolyzing voltage (MEV), the greater the energy requirement for a given electrolytic process. The MEV of a given electrochemical reaction cannot be changed; however, by changing the reaction itself, e.g., the anodic half-cell reaction, it is possible to substantially decrease the energy needed to obtain a given product.
Currently, oxygen evolution is the main anode reaction utilized in copper electrowinning. However, a number of novel reactions are receiving increasing attention. For each anode reaction listed below, determine the corresponding MEV for copper electrowinning, assuming standard conditions: